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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then f(x) is continuous at $\text{x}=\frac{\pi}{2},$ if:
  • A
    $\text{a}=\frac{1}{3},\text{ b}=2$
  • ✓
    $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
  • C
    $\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
  • D
    none of these
✓

Answer

Correct option: B.
$\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$

We have

$\Big(\text{LHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$

$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$

$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$

$=\frac{1}{3}$

$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$

$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$

$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$

$=\frac{\text{b}}{8}\times1$

$=\frac{\text{b}}{8}$

Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$

If f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\text{f}\big(\frac{\pi}{2}\big)$

$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$

$\Rightarrow\text{a}=\frac{1}{3}\text{ and }\text{b}=\frac{8}{3}$

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